- #1
Mayan Fung
- 131
- 14
The 2D Laplacian in polar coordinates has the form of
$$ \frac{1}{r}(ru_r)_r +\frac{1}{r^2}u_{\theta \theta} =0 $$
By separation of variables, we can write the ## \theta## part as
$$ \Theta'' (\theta) = \lambda \Theta (\theta)$$
Now, the book said because we need to satisfy the condition ## \Theta (0)=\Theta (2\pi)##, ##\lambda = n^2## with n =1,2,3...
I tried this in detailed steps.
##\Theta (\theta) = Acos\sqrt{\lambda}\theta + Bcos\sqrt{\lambda}\theta##
With the BC,
## B = Acos(\sqrt{\lambda}2\pi) + Bsin(\sqrt{\lambda}2\pi) ##
This is just an equality but not an identity. Why can't I conclude that
## B = \frac{cos(\sqrt{\lambda}2\pi)}{1-sin(\sqrt{\lambda}2\pi)} A ##
instead of imposing constraint on ##\lambda##
Thanks!
$$ \frac{1}{r}(ru_r)_r +\frac{1}{r^2}u_{\theta \theta} =0 $$
By separation of variables, we can write the ## \theta## part as
$$ \Theta'' (\theta) = \lambda \Theta (\theta)$$
Now, the book said because we need to satisfy the condition ## \Theta (0)=\Theta (2\pi)##, ##\lambda = n^2## with n =1,2,3...
I tried this in detailed steps.
##\Theta (\theta) = Acos\sqrt{\lambda}\theta + Bcos\sqrt{\lambda}\theta##
With the BC,
## B = Acos(\sqrt{\lambda}2\pi) + Bsin(\sqrt{\lambda}2\pi) ##
This is just an equality but not an identity. Why can't I conclude that
## B = \frac{cos(\sqrt{\lambda}2\pi)}{1-sin(\sqrt{\lambda}2\pi)} A ##
instead of imposing constraint on ##\lambda##
Thanks!